Question

Искусственный спутник земли движется по круговой орбите на расстоянии h от поверхности земли,Какова линейная скорость спутника?Уменьшится или увеличится скорость при переходе на более низкую орбиту?определите период обращения спутника

Answer 1

Дано:
h
Найти: v, T
Решение:
M₃ - масса Земли
m - масса спутника
По закону Всемирного тяготения
$F=G \frac{mM_3}{r^2}$
Радиус орбиты
$r=R+h$
где R - радиус Земли
Тогда
$F=G \frac{mM_3}{(R+h)^2}$
Эта сила сообщает спутнику центростремительное ускорение. По Второму закону Ньютона
$a= \frac{F}{m} =G \frac{M_3}{(R+h)^2}$
С другой стороны, центростремительное ускорение вычисляется по формуле
$a= \frac{v^2}{r}= \frac{v^2}{R+h}$
Следовательно
$\frac{v^2}{R+h}=G \frac{M_3}{(R+h)^2} \\ v^2=G \frac{M_3}{R+h} \\ v= \sqrt{G \frac{M_3}{R+h}}$
Анализ полученного выражения показывает, что при переходе на более низкую орбиту скорость увеличится (h меньше - знаменатель меньше - вся дробь больше)
Период находим деля путь на скорость
$T= \frac{2 \pi r}{v} = \frac{2 \pi (R+h)}{v} = \frac{2 \pi (R+h)}{\sqrt{G \frac{M_3}{R+h}}} = \\ =2 \pi \sqrt{ \frac{(R+h)^3}{GM_3} }$
Ответ:
$v= \sqrt{G \frac{M_3}{R+h}};$
Скорость увеличится;
$T=2 \pi \sqrt{ \frac{(R+h)^3}{GM_3} }$

Answer 2

Linear Velocity of a Satellite

The linear velocity of a satellite in circular orbit around the Earth can be determined using the formula:

v = √(G * M / r)

Where: - v is the linear velocity of the satellite - G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2) - M is the mass of the Earth (approximately 5.972 × 10^24 kg) - r is the distance between the satellite and the center of the Earth (equal to the sum of the radius of the Earth and the altitude of the satellite)

To calculate the linear velocity of the satellite, we need to know the altitude of the satellite from the surface of the Earth.

Effect of Lowering Orbit on Velocity

When a satellite transitions to a lower orbit, its linear velocity increases. This is because the distance between the satellite and the center of the Earth decreases, resulting in a smaller radius (r) in the velocity formula. As a result, the linear velocity of the satellite increases.

Period of Satellite Orbit

The period of a satellite's orbit is the time it takes for the satellite to complete one full revolution around the Earth. It can be calculated using the formula:

T = 2π * √(r^3 / (G * M))

Where: - T is the period of the satellite's orbit - G is the gravitational constant - M is the mass of the Earth - r is the distance between the satellite and the center of the Earth

To calculate the period of the satellite's orbit, we need to know the altitude of the satellite from the surface of the Earth.

Now, let's calculate the linear velocity and period of the satellite's orbit based on the given information.

Calculation

Given: - The satellite is in a circular orbit at a distance h from the surface of the Earth.

To calculate the linear velocity, we need to determine the distance between the satellite and the center of the Earth (r). This can be calculated by adding the radius of the Earth (R) to the altitude of the satellite (h):

r = R + h

The radius of the Earth is approximately 6,371 kilometers (or 6,371,000 meters).

Now, we can calculate the linear velocity (v) using the formula mentioned earlier:

v = √(G * M / r)

To calculate the period of the satellite's orbit (T), we can use the formula mentioned earlier:

T = 2π * √(r^3 / (G * M))

Let's calculate the linear velocity and period of the satellite's orbit based on the given information.

Calculation Example

Given: - The satellite is in a circular orbit at a distance **